Question

parallel-plate capacitor is made of two square plates 25 cm on a side and 1.0 mmapart. The capacitor is connected to a 50.0-V battery. With the battery still connected,the plates are pulled apart to a separation of 2.00 mm. What are the energies stored inthe capacitor before and after the plates are pulled farther apart? Why does the energydecrease even though work is done in separating the plates.

Answer 1

Answer:

6.9 x 10^-7 J  

3.5 x 10^-7 J

Explanation:

Identify the unknown:  

The energies stored in the capacitor before and after the plates are pulled farther apart  

List the Knowns:

Voltage of the battery: V = 50 V

Area of the plates: A = 0.25 x 0.25 = 0.0625 m^2

Original distance between the plates: d = 1 mm = 10^-3 m

New distance between the plates: d = 2 mm = 2 x 10^-3 m

Permittivity of free space: ∈o = 8.85 x 10^-12 C^2/Nm^2-

Set Up the Problem:  

Capacitance of a parallel-plate capacitor:  

C=∈o*A/d

Energy stored in a capacitor:  

U_c=(1/2)*V^2*C

Solve the Problem:  

Before the plates are pulled farther apart:  

C = 8.85 x 10^-12 x 0.0625/10^-3 = 5•53 x 10^-10 F  

U_c = (1/2) x (50)^2 x 5.53 x 10^-10= 6.9 x 10^-7 J  

After the plates are pulled farther apart:  

C = 8.85 x 10^-12 x 0.0625/2*10^-3 = 2•77 x 10^-10 F  

U_c = (1/2) x (50)^2 x 2•77 x 10^-10 = 3.5 x 10^-7 J  

The energy decrease because the capacitance decrease, so the stored charge decrease and transferred to the battery